Studied by 18 people
University of South Carolina - Steven Derocher's Math241 topics covered, focused on 13.1-13.4 and any additional formulas given
(a=alpha) horizontal component of a projectile
x0+v0cos(a)t
(a=alpha) vertical component of a projectile
y0+v0sin(a)t-1/2(g)t^2
a (alpha)=
the angle formed by the projectile
the gravitational constant, g, equals
9.8
to solve for the horizontal distance of a projectile
set the y component equal to 0, solve for t, then plug this value into the x component
(a=alpha) ideal projectile motion from point (x0,y0), r=
(x+|v0|(cosa)t)i+(y+|v0|(sina)t-(1/2)gt^2)j
v0 is the
initial velocity of the projectile
length of a smooth curve (r=x(t)i+y(t)j+z(t)k), L=
∫a→b sqrt((dx/dt)^2+(dy/dt)^2+dz/dt)^2)dt
arc length formula (simplified)
∫a→b|v(t)|dt
(T=tau) arc length parameter with base point P(t0), s(t)=
∫t0→t sqrt((x’T)^2+(y’T)^2+(z’T))dT
(T=tau) simplified arc length parameter, s(t)=
∫t0→t|v(T)|dT
unit tangent vector, T(t)=
v(t)/|v(t)|, the direction the particle is travelling in
speed on a smooth curve
ds/dt=|v(t)|
(k=kappa) curvature formula 1 (good for understanding) k=
|dT/ds|
(k=kappa, T=unit tangent vector) second curvature formula, k=
(1/|r’(t)|)(dT/dt)
principal unit normal vector, N=
(dT/dt)/(|dT/dt|), the direction T(t) is changing in
(k=kappa) third curvature formula (the one we typically use), k=
(|T(t)|X|N(t)|)/(|T(t)|)^3
radius of an osculating circle
1/k
bynormal vector, B=
TXN
TNB frame
k of a circle is…, k of a line is…
constant, 0
r(t) is orthogonal to r’(t) if
r(t) is differentiable and |r(t)| is constant
(d/dt)(u(f(t)))=f’(t)(u’(f(t))
(projectile motion) to find distance travelled by projectile,…
set j component equal to 0, solve for t (throw out any t’s that don’t make sense), plug t into r(t), solve
to find max height of projectile,…
solve for t where r’(t)=0, check that it is a max using 2nd derivative test (left side should be positive, right side should be negative), plug this t into r(t)
Note
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